To determine the number of ways to choose 6 members from a chorus of 30 for a special performance, we use the concept of combinations, as we are selecting a subset where the order does not matter.
The number of combinations is given by the binomial coefficient, often denoted as [tex]\( \binom{n}{k} \)[/tex], where [tex]\( n \)[/tex] is the total number of items to choose from, and [tex]\( k \)[/tex] is the number of items to choose. The formula for combinations is:
[tex]\[
\binom{n}{k} = \frac{n!}{k!(n-k)!}
\][/tex]
In this problem, [tex]\( n = 30 \)[/tex] and [tex]\( k = 6 \)[/tex], so the formula becomes:
[tex]\[
\binom{30}{6} = \frac{30!}{6! \cdot (30-6)!} = \frac{30!}{6! \cdot 24!}
\][/tex]
Let's examine the expressions given:
1. [tex]\( 30 \cdot 6 \)[/tex]:
- This represents multiplying 30 by 6, which is not the correct method for calculating combinations.
2. [tex]\( \frac{30!}{24!} \)[/tex]:
- This expression indicates dividing the factorial of 30 by the factorial of 24. This does not account for the necessity of also dividing by the factorial of the group size (6 in this case) in combinations.
3. [tex]\( \frac{30!}{6!24!} \)[/tex]:
- This is the correct expression. It correctly uses the combination formula by dividing the factorial of the total number of items (30) by the product of the factorial of the group size (6) and the factorial of the remainder (24).
Thus, the correct expression that represents the number of ways to choose a group of 6 members from 30 for a special performance is:
[tex]\[
\frac{30!}{6! \cdot 24!}
\][/tex]
Using this expression, the numerical result for the number of ways to choose the group is found to be:
[tex]\[
593,775
\][/tex]
So, the correct option is: [tex]\(\frac{30!}{6!24!}\)[/tex].
